Question

For what values of c are the following systems inconsistent, with unique solution or with infinitely many solutions? (1) cx1 + x2 + x3 =2; x1 +cx2+ x3 = 2; x1 + x2 + cx3 = 2

Answer 1

\[\begin{pmatrix}c&1&1\\1&c&1\\1&1&c\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}2\\2\\2\end{pmatrix}\]
Using Cramer's rule, we would find solutions
\[\begin{matrix}x_1=\dfrac{\begin{vmatrix}2&1&1\\2&c&1\\2&1&c\end{vmatrix}}{\begin{vmatrix}c&1&1\\1&c&1\\1&1&c\end{vmatrix}}&&
x_2=\dfrac{\begin{vmatrix}c&2&1\\1&2&1\\1&2&c\end{vmatrix}}{\begin{vmatrix}c&1&1\\1&c&1\\1&1&c\end{vmatrix}}&&
x_3=\dfrac{\begin{vmatrix}c&1&2\\1&c&2\\1&1&2\end{vmatrix}}{\begin{vmatrix}c&1&1\\1&c&1\\1&1&c\end{vmatrix}}\end{matrix}\]
The system is inconsistent if the denominator determinant is zero, and it has multiple solutions if both the numerators and denominator are zero.

Answer 2

Compute the determinant of the deominator first (using a cofactor expansion along the first row):
\[\begin{align*}\begin{vmatrix}c&1&1\\1&c&1\\1&1&c\end{vmatrix}&=c\begin{vmatrix}c&1\\1&c\end{vmatrix}-\begin{vmatrix}1&1\\1&c\end{vmatrix}+\begin{vmatrix}1&c\\1&1\end{vmatrix}\\\\&=c(c^2-1)-(c-1)+(c-1)\\\\&=c(c-1)(c+1)\end{align*}\]
What values of \(c\) make this zero?

Answer 3

Whoops, slight typo:\[\begin{align*}\begin{vmatrix}c&1&1\\1&c&1\\1&1&c\end{vmatrix}&=c\begin{vmatrix}c&1\\1&c\end{vmatrix}-\begin{vmatrix}1&1\\1&c\end{vmatrix}+\begin{vmatrix}1&c\\1&1\end{vmatrix}\\\\&=c(c^2-1)-(c-1)+\color{red}{(1-c)}\\\\&=\cdots\end{align*}\]